LRU cache in Java with Generics and O(1) operations
From the question itself, we can see that the problem of O(n) operations arises when querying the linked list. Therefore, we need an alternative data structure. We need to be able to update the items' last access time from the HashMap without searching.
We can keep two separate data structures. A HashMap with (Key,Pointer) pairs and a doubly linked list which will work as the priority queue for deletion and store the Values. From the HashMap, we can point to an element in the doubly linked list and update its' retrieval time. Because we go directly from the HashMap to the item in the list, our time complexity remains at O(1)
For example, our doubly linked list can look like:
least_recently_used -> A <-> B <-> C <-> D <-> E <- most_recently_used
We need to keep a pointer to the LRU and MRU items. The entries' values will be stored in the list and when we query the HashMap, we will get a pointer to the list. On get(), we need to put the item at the right-most side of the list. On put(key,value), if the cache is full, we need to remove the item at the left-most side of the list from both, the list and the HashMap.
The following is an example implementation in Java:
public class LRUCache<K, V>{
// Define Node with pointers to the previous and next items and a key, value pair
class Node<T, U> {
Node<T, U> previous;
Node<T, U> next;
T key;
U value;
public Node(Node<T, U> previous, Node<T, U> next, T key, U value){
this.previous = previous;
this.next = next;
this.key = key;
this.value = value;
}
}
private HashMap<K, Node<K, V>> cache;
private Node<K, V> leastRecentlyUsed;
private Node<K, V> mostRecentlyUsed;
private int maxSize;
private int currentSize;
public LRUCache(int maxSize){
this.maxSize = maxSize;
this.currentSize = 0;
leastRecentlyUsed = new Node<K, V>(null, null, null, null);
mostRecentlyUsed = leastRecentlyUsed;
cache = new HashMap<K, Node<K, V>>();
}
public V get(K key){
Node<K, V> tempNode = cache.get(key);
if (tempNode == null){
return null;
}
// If MRU leave the list as it is
else if (tempNode.key == mostRecentlyUsed.key){
return mostRecentlyUsed.value;
}
// Get the next and previous nodes
Node<K, V> nextNode = tempNode.next;
Node<K, V> previousNode = tempNode.previous;
// If at the left-most, we update LRU
if (tempNode.key == leastRecentlyUsed.key){
nextNode.previous = null;
leastRecentlyUsed = nextNode;
}
// If we are in the middle, we need to update the items before and after our item
else if (tempNode.key != mostRecentlyUsed.key){
previousNode.next = nextNode;
nextNode.previous = previousNode;
}
// Finally move our item to the MRU
tempNode.previous = mostRecentlyUsed;
mostRecentlyUsed.next = tempNode;
mostRecentlyUsed = tempNode;
mostRecentlyUsed.next = null;
return tempNode.value;
}
public void put(K key, V value){
if (cache.containsKey(key)){
return;
}
// Put the new node at the right-most end of the linked-list
Node<K, V> myNode = new Node<K, V>(mostRecentlyUsed, null, key, value);
mostRecentlyUsed.next = myNode;
cache.put(key, myNode);
mostRecentlyUsed = myNode;
// Delete the left-most entry and update the LRU pointer
if (currentSize == maxSize){
cache.remove(leastRecentlyUsed.key);
leastRecentlyUsed = leastRecentlyUsed.next;
leastRecentlyUsed.previous = null;
}
// Update cache size, for the first added entry update the LRU pointer
else if (currentSize < maxSize){
if (currentSize == 0){
leastRecentlyUsed = myNode;
}
currentSize++;
}
}
}
Implementation that passes the tests of the leetcode questiton with simple unit tests
I have made a pull request with this at: https://github.com/haoel/leetcode/pull/90/files
LRUCache.java
import java.util.Iterator;
import java.util.LinkedHashMap;
public class LRUCache {
private int capacity;
private LinkedHashMap<Integer,Integer> map;
public LRUCache(int capacity) {
this.capacity = capacity;
this.map = new LinkedHashMap<>(16, 0.75f, true);
}
public int get(int key) {
Integer value = this.map.get(key);
if (value == null) {
value = -1;
}
return value;
}
public void put(int key, int value) {
if (
!this.map.containsKey(key) &&
this.map.size() == this.capacity
) {
Iterator<Integer> it = this.map.keySet().iterator();
it.next();
it.remove();
}
this.map.put(key, value);
}
}
LRUCacheTest.java
public class LRUCacheTest {
public static void main(String[] args) {
LRUCache c;
// Starts empty.
c = new LRUCache(2);
assert c.get(1) == -1;
// Below capcity.
c = new LRUCache(2);
c.put(1, 1);
assert c.get(1) == 1;
assert c.get(2) == -1;
c.put(2, 4);
assert c.get(1) == 1;
assert c.get(2) == 4;
// Above capacity, oldest is removed.
c = new LRUCache(2);
c.put(1, 1);
c.put(2, 4);
c.put(3, 9);
assert c.get(1) == -1;
assert c.get(2) == 4;
assert c.get(3) == 9;
// get renews entry
c = new LRUCache(2);
c.put(1, 1);
c.put(2, 4);
assert c.get(1) == 1;
c.put(3, 9);
assert c.get(1) == 1;
assert c.get(2) == -1;
assert c.get(3) == 9;
// Double put does not remove due to capacity.
c = new LRUCache(2);
assert c.get(2) == -1;
c.put(2, 6);
assert c.get(1) == -1;
c.put(1, 5);
c.put(1, 2);
assert c.get(1) == 2;
assert c.get(2) == 6;
}
}
removeEldestEntry() alternative implementation
Not sure it is worth it as it takes the same number of lines, but here goes for completeness:
import java.util.LinkedHashMap;
import java.util.Iterator;
import java.util.Map;
import java.io.*;
class LinkedhashMapWithCapacity<K,V> extends LinkedHashMap<K,V> {
private int capacity;
public LinkedhashMapWithCapacity(int capacity) {
super(16, 0.75f, true);
this.capacity = capacity;
}
@Override
protected boolean removeEldestEntry(Map.Entry<K,V> eldest) {
return this.size() > this.capacity;
}
}
public class LRUCache {
private LinkedhashMapWithCapacity<Integer,Integer> map;
public LRUCache(int capacity) {
this.map = new LinkedhashMapWithCapacity<>(capacity);
}
public int get(int key) {
Integer value = this.map.get(key);
if (value == null) {
value = -1;
}
return value;
}
public void put(int key, int value) {
this.map.put(key, value);
}
}
Tested on Ubuntu 20.10, OpenJDK 11.0.10.
The LinkedHashMap designed with that in mind
From the javadocs:
A special constructor is provided to create a linked hash map whose order of iteration is the order in which its entries were last accessed, from least-recently accessed to most-recently (access-order). This kind of map is well-suited to building LRU caches. Invoking the put, putIfAbsent, get, getOrDefault, compute, computeIfAbsent, computeIfPresent, or merge methods results in an access to the corresponding entry (assuming it exists after the invocation completes). The replace methods only result in an access of the entry if the value is replaced. The putAll method generates one entry access for each mapping in the specified map, in the order that key-value mappings are provided by the specified map's entry set iterator. No other methods generate entry accesses. In particular, operations on collection-views do not affect the order of iteration of the backing map.
The removeEldestEntry(Map.Entry) method may be overridden to impose a policy for removing stale mappings automatically when new mappings are added to the map.